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It is a direct consequence of Fubini's theorem that if $f,g \in L^1(\mathbb{R})$, then the convolution $f *g$ is well defined almost everywhere and $f*g \in L^1(\mathbb{R})$. Thus, $L^1(\mathbb{R})$ is closed under convolution, and it is a Banach algebra without unit since we have the inequality

$$\|f*g\|_{1} \leq \|f\|_1 \|g\|_1 \qquad (f,g \in L^1(\mathbb{R})).$$ Now, it follows from Hölder's inequality that if $f,g \in L^2(\mathbb{R})$, then $f*g$ is bounded.

My question is the following : Does $f*g$ necessarily belongs to $L^2(\mathbb{R})$? In other words, is $L^2(\mathbb{R})$ closed under convolution?

Since a quick google search seem to result in a negative answer, I also ask the following question :

Can you give an explicit example of two functions $f, g \in L^2(\mathbb{R})$ such that $f*g \notin L^2(\mathbb{R})$?

Thank you, Malik

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Have you tried the ideas in the answers to this question? – user21467 Mar 9 '12 at 15:15
@Steven Taschuk : Could you please indicate what answers you're refering to? The accepted answer shows that there is no inequality $\|f∗g\|_2 \leq \|f\|_2\|g\|_2$, assuming $L^2$ is closed under convolution. As for the other answers, perhaps I am mistaken, but I don't see how those Fourier arguments could work here : the identity $\hat{f*g} = \hat{f}\hat{g}$ is valid for$f,g$ in $L^1$... – Kalim Mar 9 '12 at 20:13
If $f\in L^1\cap L^2$, then $\int|f|^2=\int|\hat f|^2$. This allows us to extend the Fourier transform as an isometry of $L^2$, and by interpolation, as a bounded operator from $L^p$ to $L^{p'}$, $1<p<2$, $1/p+1/p'=1$. Thus, the answers in Steven's link do answer your question. An explicit example is $f(x)=g(x)={}_1F_2(\frac38;\frac12,\frac{11}{8};-\frac{x^2}{2})$. – Julián Aguirre Mar 9 '12 at 22:44
@JuliánAguirre I don't understand your notation, could you write as a an answer what exactly are the functions $f$ and $g$ that work? – Kalim Mar 10 '12 at 0:00
Regarding his notation: – Chris Janjigian Mar 10 '12 at 0:19
up vote 7 down vote accepted

Since the Fourier Transform of the product of two functions is the same as the convolution of their Fourier Transforms, and the Fourier Transform is an isometry on $L^2$, all we need find is an $L^2$ function that when squared is no longer an $L^2$ function. Take the function $$ f(x)=e^{-x^2}|x|^{-1/3} $$ $f\in L^2(\mathbb{R})$, yet $f^2\not\in L^2(\mathbb{R})$. Thus, $\hat{f}\in L^2(\mathbb{R})$, yet $\hat{f}*\hat{f}\not\in L^2(\mathbb{R})$.


The reason that it is hard to come up with an explicit example without using the Fourier Transform, is that the $L^2$ functions involved in the convolution do not decay at $\infty$ quickly enough to be integrable; that is, the convolution requires cancellation to evaluate. The $\hat{f}$ given above is not in $L^1$ (if it were, then $f$ would be bounded), so trying to compute the convolution with itself would be extremely difficult.

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Thank you very much for your answer. Just wondering though : when you say that the fourier transform of the product of two functions is the convolution of their Fourier transforms (known as the convolution theorem), is it required that those two functions are in $L^1$? The proof I know uses the fact that the functions are integrable... Does it also work if the functions are only assumed to be in $L^2$? Note that your function $f$ is in $L^1$, so everything is fine, but this is what troubled me when reading the answers to the other question. – Kalim Mar 10 '12 at 14:30
@Malik: The Schwartz Class, $\mathscr{S}$ is dense in $L^2$ and each function in $\mathscr{S}$ is in $L^1$. So we define the Fourier Transform of a general $f\in L^2$ by finding a sequence $f_n\in\mathscr{S}$ so that $f_n\to f$ in $L^2$. The fact that $\|f\|_{L^2}=\|\hat{f}\|_{L^2}$ shows that $\widehat{f_n}\to\hat{f}$. – robjohn Mar 10 '12 at 14:37

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